We want to find the probability that a student who attends the football game also attends the dance. To do so, we will define two events.

$A :$ a student attends the football game
$B :$ a student attends the dance

The probability that an event will occur, given that another event has already occurred, is called a conditional probability. Therefore, we are interested in

P (attends the dance ∣ attends the game),

which can be denoted in the following way.

P (B ∣ A)

We will rewrite the formula for the probability of dependent events to have a rule for finding the desired conditional probability.

P (A and B) = P (A) \cdot P (B ∣ A)

DivEqn

$LHS / P (A) = RHS / P (A)$

\frac{P ( A and B )}{P ( A )} = P (B ∣ A)

RearrangeEqn

Rearrange equation

P (B ∣ A) = \frac{P ( A and B )}{P ( A )}

We know that

43 %

of students attend the football game, which corresponds to the probability of event

A .

P (A) = 43 % \Leftrightarrow P (A) = 0.43

We are also told that

23 %

of students go both to the game and to the dance. Therefore, we can write the probability

P (A and B) .

P (A and B) = 23 % \Leftrightarrow P (A and B) = 0.23

With this information, we can calculate the

P (B ∣ A) .

P (B ∣ A) = \frac{P ( A and B )}{P ( A )}

SubstituteII

$P (A and B) = 0.23$ , $P (A) = 0.43$

P (B ∣ A) = \frac{0 . 2 3}{0 . 4 3}

CalcQuot

Calculate quotient

P (B ∣ A) = 0.534883 \dots

RoundDec

Round to $3$ decimal place(s)

P (B ∣ A) \approx 0.535

WritePercent

Convert to percent

P (B ∣ A) = 53.5 %

The probability that a student who attended the homecoming football game also attended the game is about

53.5 % .

Exercise 23 Page 681

Hints

Check the answer

Practice exercises

Asses your skill